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During a rockslide, a \(524-\mathrm{kg}\) rock slides from rest down a hill slope that is \(488 \mathrm{~m}\) long and \(292 \mathrm{~m}\) high. The speed of the rock as it reaches the bottom of the hill is \(62.6 \mathrm{~m} / \mathrm{s}\). How much mechanical energy does the rock lose in the slide due to friction?

Short Answer

Expert verified
The rock loses \(4.7 * 10^6 J\) of mechanical energy due to friction.

Step by step solution

01

Calculate initial potential energy

Calculate the initial potential energy that the rock has at the top of the hill using the equation \(PE_i = mgh\), where \(m\) is the mass of the rock, \(g\) is the acceleration due to gravity (approximated as \(9.8 m/s^2\)), and \(h\) is the height of the hill. Plugging the given values this results in \(PE_i = 524 kg * 9.8 m/s^2 * 292 m = 1.5 * 10^7 J\).
02

Calculate final kinetic energy

Calculate the final kinetic energy that the rock has at the bottom of the hill using the equation \(KE_f = \frac{1}{2}mv^2\), where \(m\) is the mass of the rock and \(v\) is the speed. Plugging the given values this results in \(KE_f = \frac{1}{2} * 524 kg * (62.6 m/s)^2 = 1.03 * 10^7 J\).
03

Calculate energy loss due to friction

The mechanical energy lost due to friction is the difference between the initial potential energy and the final kinetic energy. Therefore, the energy lost due to friction is calculated as \(E_f = PE_i - KE_f = 1.5 * 10^7 J - 1.03 * 10^7 J = 4.7 * 10^6 J\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When considering an object sliding down a hill, like the rock in our exercise, its kinetic energy increases as it accelerates downward. This kind of energy is quantified using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \(m\) is the mass of the object and \(v\) is its velocity.

In this context, as the rock descends the hill, it transitions from having no kinetic energy (since it starts from rest) to a significant amount as it reaches the bottom with a speed of 62.6 m/s. This is calculated in our exercise as 1.03 \(\times\) 107 Joules at the hill's base.

Understanding kinetic energy helps us grasp how energy changes forms as the rock moves, showcasing the dynamic transformation within mechanical systems.
Potential Energy
Potential energy is the stored energy in an object due to its position or height. For a rock perched atop a hill, potential energy depends on how high it is positioned above the ground, and it’s calculated using the formula:
  • \( PE = mgh \)
where \(m\) is the mass, \(g\) is the gravitational acceleration (approximated as 9.8 m/s2), and \(h\) is the height.

The rock in our exercise has an initial potential energy since it starts at rest at the top of a 292-meter high hill. This gives it a potential energy of 1.5 \(\times\) 107 Joules. As the rock slides down, this potential energy is converted into kinetic energy.

Recognizing potential energy is crucial for understanding energy conservation as it highlights how energy is stored and then transformed during movements like sliding down a slope.
Energy Loss Due to Friction
During energy transformations, not all energy is converted perfectly, and some are lost due to frictional forces. Friction is a force that opposes the relative motion of surfaces, converting mechanical energy into heat, causing an energy loss.In the rockslide problem, we can calculate the energy lost to friction by comparing the initial potential energy and the final kinetic energy:
  • \( E_{loss} = PE_i - KE_f \)
Plugging in our numbers, the initial potential energy was 1.5 \(\times\) 107 Joules, while the final kinetic energy was 1.03 \(\times\) 107 Joules. The difference, 4.7 \(\times\) 106 Joules, represents the energy lost to friction.

Understanding energy loss is important for appreciating realistic scenarios, where systems are not perfectly efficient, emphasizing the impact of friction in mechanical processes.

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Most popular questions from this chapter

The National Transportation Safety Board is testing the crashworthiness of a new car. The \(2340-\mathrm{kg}\) vehicle is driven at \(12.6 \mathrm{~km} / \mathrm{h}\) into an abutment. During impact, the center of mass of the car moves forward \(64.0 \mathrm{~cm} ;\) the abutment is compressed by \(8.30 \mathrm{~cm}\). Ignore friction between the car and the road. (a) Find the force, assumed constant, exerted by the abutment on the car. (b) By how much does the internal energy of the car increase?

While a \(1700-\mathrm{kg}\) automobile is moving at a constant speed of \(15 \mathrm{~m} / \mathrm{s}\), the motor supplies \(16 \mathrm{~kW}\) of power to overcome friction, wind resistance, and so on. (a) What power must the motor supply if the car is to move up an \(8.0 \%\) grade \((8.0 \mathrm{~m}\) vertically for each \(100 \mathrm{~m}\) horizontally) at \(15 \mathrm{~m} / \mathrm{s} ?\) (b) At what downgrade, expressed in percentage terms, would the car coast at \(15 \mathrm{~m} / \mathrm{s}\) ?

An automobile with passengers has a weight of \(16,400 \mathrm{~N}\) \((=3680 \mathrm{lb})\) and is moving up a \(10^{\circ}\) slope with an initial speed of \(70 \mathrm{mi} / \mathrm{h}(=113 \mathrm{~km} / \mathrm{h})\) when the driver begins to apply the brakes. The car comes to a stop after traveling \(225 \mathrm{~m}\) along the inclined road. Calculate the work done by the brakes in stopping the car, assuming that all other energy transfers in this problem (such as heat and internal energy) can be neglected.-

A rubber ball dropped from a height of exactly \(6 \mathrm{ft}\) bounces (hits the floor) several times, losing \(10 \%\) of its kinetic energy each bounce. After how many bounces will the ball subsequently not rise above \(3 \mathrm{ft}\) ?

A \(25.3-\mathrm{kg}\) bear slides, from rest, \(12.2 \mathrm{~m}\) down a lodge- pole pine tree, moving with a speed of \(5.56 \mathrm{~m} / \mathrm{s}\) at the bottom. \((a)\) What is the initial potential energy of the bear? \((b)\) Find the kinetic energy of the bear at the bottom. (c) Assuming no other energy transfers, find the change in internal energy of the bear and tree.

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